Walk me through this proof that a theory is satisfiable.

Question

Setting

Suppose $\mathcal M, \mathcal N \models T$, $\mathcal M \subseteq \mathcal N$, $\mathcal M$ existentially closed, then I I want to prove that there is $\mathcal M_1 \models T$ so that $\mathcal M \subseteq \mathcal N \subseteq \mathcal M_1$ with $\mathcal M \prec \mathcal M_1$.

Solution

Create $\mathcal L_N$ theory $T' = Diag_{el}(\mathcal M) \cup Diag(\mathcal N)$ and show that T' is satisfiable. If so let $\mathcal M_1 \models T'$.

Let $\triangle \subseteq T'$, $\triangle$ some arbitrary finite subset, and by compactness if all such $\triangle$'s are satisfiable, then $T'$ is satisfiable. Now I am having trouble proving how $\triangle$ is satisfiable.

Suppose $\triangle = \Gamma \cup \Sigma$ where $\Gamma = \{\phi : \phi \in Diag_{el} \mathcal M \cap \triangle\}$ and $\Sigma = \{\psi : \psi \in Diag(\mathcal N) \cap \triangle\}$, then if $\triangle$ is inconsistent, we have for $\bar{b} \in \mathbb N, \bar{a} \in \mathbb M$:

$$\Gamma \models \neg \Big(\bigwedge \psi(\bar{b},\bar{a})\Big)\\ \Gamma \models \bigvee \neg \psi(\bar{b},\bar{a})\\ \Gamma \models \bigvee \neg \exists \bar x \psi(\bar{x},\bar{a})\\ \Gamma \models \neg (\exists \bar x \psi(\bar{x},\bar{a})).$$

But since $\mathcal M$ is existentially closed, we know whenver $\mathcal N \models \exists \bar x\psi(\bar{x},\bar{a})$, $\mathcal M \models \exists \bar x \psi(\bar x, \bar a)$, a contradiction.

So my problems is that I am not sure if the first line in the statement is correct, in particular $\psi$ is a sentence from $Diag(\mathcal N)$ but could have parameters in $\mathbb M$? And feel free to point out any other wrong statements in my argument.

Answer 1

A few scattered comments.

1) You need to ensure that $\bar{b} \in \mathbb{N} - \mathbb{M}$ in order for your step from $\Gamma \models \bigvee \neg \psi(\bar{b}, \bar{a})$ to $\Gamma \models \bigvee \neg \exists \bar{x} \psi(\bar{x}, \bar{a})$ to work. This is simply a matter of rewriting the members of $\Sigma$ -- a technicality at best.

2.a) The step from $\Gamma \models \bigvee \neg \exists \bar{x} \psi(\bar{x}, \bar{a})$ to $\Gamma \models \neg \exists \bar{x} \psi(\bar{x}, \bar{a})$ does not make any sense because of the $\psi$ in the last expression. What is $\psi$ here?

2.b) What you really want to do is exclude the last step that contains the mysterious $\psi$ and derive the contradiction directly from $\Gamma \models \bigvee \neg \exists \bar{x} \psi(\bar{x}, \bar{a})$. Since $\mathcal{M} \models \Gamma$, you have derived that $\mathcal{M} \models \bigvee \neg \exists \bar{x} \psi(\bar{x}, \bar{a})$. But for each such $\psi \in \Sigma$ we have that $\mathcal{N} \models \exists \bar{x} \psi(\bar{x}, \bar{a})$, and hence as $\mathcal{M}$ is existentially closed in $\mathcal{N}$ we have that $\mathcal{M} \models \exists \bar{x} \psi(\bar{x}, \bar{a})$ as well. Therefore $\mathcal{M} \models \neg (\bigvee \neg \exists \bar{x} \psi(\bar{x}, \bar{a}))$, a contradiction. You wrote this part in your argument.

3) In my opinion it is best not to prove the satisfiability of $\Delta$ by contradiction but rather prove it directly. Do this as follows. Simply write $\Delta = \{\phi_1(\bar{m}), \ldots, \phi_t(\bar{m}), \psi_1(\bar{n}), \ldots, \psi_u(\bar{n})\}$ wherein the $\phi_i$ are in $Diag_{el}(\mathcal{M})$ and the $\psi_j$ are in $Diag(\mathcal{N})$, as you have written. Now simply replace the $\bar{n}$ by elements of $\mathcal{M}$ by using the fact that $\mathcal{M}$ is existentially closed in $\mathcal{N}$. This gives you a way of satisfying $\Delta$ using $\mathcal{M}$.